Question

我有以下代码，在登录功能中，输出是错误的（逻辑错误）。它基本上打印“无效的用户名和密码”，直到它到达正确的用户名和密码，然后打印“正确登录”。

ERRONEOUS OUTPUT ：

例如，使用测试数据：user3和pass3，输出为：

*****LOGIN SCREEN******
Username: user3
Password: pass3
invalid username or password
invalid username or password
correct login
>>>

这是代码，参考LOGIN函数：

usernames=["user1","user2","user3"]
passwords=["pass1","pass2","pass3"]

def main():
   mainmenu()


def mainmenu():
   print("****MAIN MENU****")
   print("=======Press L to login :")
   print("=======Press R to register :")
   choice1=input()
   if choice1=="L" or choice1=="l":
      login()
   elif choice1=="R" or choice1=="r":
      register()
   else:
      print("please make a valid selection")

usernames=["user1","user2","user3"]
passwords=["pass1","pass2","pass3"]

def login():
  print("*****LOGIN SCREEN******")
  username=input("Username: ")
  password=input("Password: ")
  for index_of_current_user, current_user in enumerate(usernames): #enumerate allows to you to go throw the list and gives to you the current element, and the index of the current element
    if username == current_user and passwords[index_of_current_user] == password: #since the two list are linked, you can use the index of the user to get the password in the passwords list
      print("correct login")
    else:
      print("invalid username or password")

def register():
  print("*****REGISTRATION****")
  username=input("Enter a username:")
  password=input("Enter a password:")
  users_pass[username] = password
  answer=input("Do you want to make another registration?")
  if answer=="y":
    register()
  else:
    registration_details()

def registration_details():
   print(usernames)
   print(passwords)

main()

我在找 a）解决问题的解决方案并修复，以便在找到正确的用户名和密码对时打印“正确登录”，而不是循环并打印每个用户名

b）解释关于为什么缩进或其他错误在这里不起作用 - 因为逻辑（用VB.Net这样的语言）是合理的。

如果没有额外的代码（和一个非常简单的修复）可以修复问题，我更喜欢它，但也许最优雅的解决方案是标志。也就是说，如果问题不仅仅是缩进或类似于我错过的东西。

Answer 1

您可以利用异常以这种方式写入登录。

def login():
  print("*****LOGIN SCREEN******")
  username=input("Username: ")
  password=input("Password: ")
  try:
    index = usernames.index(username)
    if password == passwords[index]:
      print("correct login")
    else:
      print("invalid username or password")
  except:
    print("invalid username or password")

Answer 2

由于登录功能中的else语句，您会发现这种情况。基本上你的代码当前在该函数中所做的是循环，检查用户名和密码是否等于当前值（即比较user1 == user2）如果它们不相等则会自动打印无效的用户名。

相反，您应该等到比较所有值以打印无效的用户名或密码消息。另外 - 一旦找到有效值，您可以添加一个中断来停止for循环，而不是继续遍历您的值。您的登录功能如下所示：

def login():
  print("*****LOGIN SCREEN******")
  username=input("Username: ")
  password=input("Password: ")
  found = False
  for index_of_current_user, current_user in enumerate(usernames): #enumerate allows to you to go throw the list and gives to you the current element, and the index of the current element
    if username == current_user and passwords[index_of_current_user] == password: #since the two list are linked, you can use the index of the user to get the password in the passwords list
      print("correct login")
      found = True
      break
  if found == False:
    print("invalid username or password")

这将只为您提供1个正确登录或无效用户名或密码的实例。

Answer 3

在你的login（）函数中，你打印列表的每个元素，所以你可以在循环后打印：

def login():
  print("*****LOGIN SCREEN******")
  username=input("Username: ")
  password=input("Password: ")
  correct_login = False
  for index_of_current_user, current_user in enumerate(usernames): #enumerate allows to you to go throw the list and gives to you the current element, and the index of the current element
    if username == current_user and passwords[index_of_current_user] == password: #since the two list are linked, you can use the index of the user to get the password in the passwords list
      correct_login = True
      break
  if(correct_login):
    print("correct login")
  else:
    print("invalid user name or password")

for循环中嵌套if语句中的特定缩进错误

3 个答案: